Hi
I did not read them all but most of them at least can be proved by induction
Hello, Sea!
I tried problem (e) by Induction.
. . It took more work than I anticipated . . .
We have .
Verify . . . . True!
Assume
Add . to both sides:
. .
. .
. . . . . . . .
. . . . . . . . . . . .
. . . . . . . . . . . .
The inductive proof is compete.
We must show
We can write f(n) in the form
where M(7) means 'is a multiple of 7'
The first term = 0 (mod 7) so now we are reduced to finding
This can be written
and this completes the proof. We have shown
i.e., the expression on the left is a multiple of 7
and
Thanks Anthony......
This is falsef)
and
I think...
This is true...
Now if n=2 we get 25 + 2 x 3 + 1 = 32 = M(8)
and if n=3 we get 125 + 2 x 9 + 1 = 144 = M(8)
and so we must prove
It will be more convenient to use f(n+1) but this does not alter the proof
We can ignore the first term as it will be divisible by 8 and look at
if n is even .......(1)
and if n is odd .......(2)
For (1) we have and putting n = 2m to make it even
= M(8) + 8 = 0 (mod 8) so OK if n is even
For (2) we have n odd and write it 2m+1 so we have
so OK if n is odd
and so for n either even or odd we have shown
and
Thanks Anthony......